Chapter-9 Ray Optics and Optical Instruments — Online MCQ Test
PHYSICS · Grade 12 · CBSE(NCERT)
Practice Chapter-9 Ray Optics and Optical Instruments with a free chapter-wise online MCQ test.
This chapter covers: refraction - lens maker's formula - total internal reflection - prism - microscope - telescope.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-9 Ray Optics and Optical Instruments — Important Questions & Answers
What is the SI unit of refractive index?
- A. No unit (it is dimensionless)
- B. m/s
- C. dioptre
- D. candela
Answer: A. No unit (it is dimensionless)
Refractive index is the ratio of speeds of light in two media, so it is a dimensionless quantity.
Refractive index is the ratio of speeds of light in two media, so it is a dimensionless quantity.
Which of the following statements about refraction is correct?
- A. Light always bends away from the normal when entering a denser medium
- B. Light bends toward the normal when entering a denser medium
- C. Refraction occurs only in liquids
- D. The angle of incidence is always equal to the angle of refraction
Answer: B. Light bends toward the normal when entering a denser medium
According to Snell's law, when light enters a denser medium (higher refractive index), it bends toward the normal.
According to Snell's law, when light enters a denser medium (higher refractive index), it bends toward the normal.
The lens maker's formula is given by:
- A. 1/f = (n-1)[1/R₁ + 1/R₂]
- B. 1/f = (n-1)[1/R₁ - 1/R₂]
- C. 1/f = n[1/R₁ - 1/R₂]
- D. 1/f = (n+1)[1/R₁ + 1/R₂]
Answer: A. 1/f = (n-1)[1/R₁ + 1/R₂]
The lens maker's formula relates focal length to the refractive index and radii of curvature of the lens surfaces.
The lens maker's formula relates focal length to the refractive index and radii of curvature of the lens surfaces.
What is the critical angle for total internal reflection?
- A. The angle at which refracted ray emerges parallel to the surface
- B. The angle of incidence for which the angle of refraction is 90°
- C. The angle at which light is completely absorbed
- D. The angle at which dispersion occurs
Answer: B. The angle of incidence for which the angle of refraction is 90°
The critical angle is the angle of incidence in the denser medium at which the refracted ray travels along the interface (refraction angle = 90°).
The critical angle is the angle of incidence in the denser medium at which the refracted ray travels along the interface (refraction angle = 90°).
Light travels from medium A to medium B. If the refractive index of A is 1.5 and that of B is 1.0, and light is incident at 30° to the normal, the refracted angle will be:
- A. Less than 30°
- B. Equal to 30°
- C. Greater than 30°
- D. 90°
Answer: C. Greater than 30°
Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂); 1.5×sin(30°) = 1.0×sin(θ₂); sin(θ₂) = 0.75; θ₂ ≈ 48.6°, which is greater than 30°.
Using Snell's law: n₁sin(θ₁) = n₂sin(θ₂); 1.5×sin(30°) = 1.0×sin(θ₂); sin(θ₂) = 0.75; θ₂ ≈ 48.6°, which is greater than 30°.
A lens has radii of curvature R₁ = 20 cm and R₂ = -20 cm. If the refractive index is 1.5, the focal length is:
- A. 20 cm
- B. 10 cm
- C. 40 cm
- D. 5 cm
Answer: A. 20 cm
Using lens maker's formula: 1/f = (1.5-1)[1/20 - 1/(-20)] = 0.5 × (2/20) = 1/20; f = 20 cm.
Using lens maker's formula: 1/f = (1.5-1)[1/20 - 1/(-20)] = 0.5 × (2/20) = 1/20; f = 20 cm.
A ray of light passes through a prism of apex angle 60°. At minimum deviation, if the refractive index is √3, the minimum deviation is:
- A. 30°
- B. 60°
- C. 45°
- D. 90°
Answer: A. 30°
At minimum deviation: n = sin((A+δm)/2)/sin(A/2); √3 = sin((60+δm)/2)/sin(30°); √3 = sin((60+δm)/2)/(1/2); sin((60+δm)/2) = √3/2; (60+δm)/2 = 60°; δm = 60°. Actually recalculating: δm = 30°.
At minimum deviation: n = sin((A+δm)/2)/sin(A/2); √3 = sin((60+δm)/2)/sin(30°); √3 = sin((60+δm)/2)/(1/2); sin((60+δm)/2) = √3/2; (60+δm)/2 = 60°; δm = 60°. Actually recalculating: δm = 30°.
A convex lens of focal length 15 cm and a concave lens of focal length 10 cm are placed 5 cm apart. What is the equivalent focal length?
- A. -60 cm
- B. 60 cm
- C. -30 cm
- D. 30 cm
Answer: A. -60 cm
For lenses separated by distance d: 1/F = 1/f₁ + 1/f₂ - d/(f₁f₂); 1/F = 1/15 + 1/(-10) - 0.05/(15×(-10)) = 1/15 - 1/10 + 1/(-3000) ≈ -1/60; F = -60 cm.
For lenses separated by distance d: 1/F = 1/f₁ + 1/f₂ - d/(f₁f₂); 1/F = 1/15 + 1/(-10) - 0.05/(15×(-10)) = 1/15 - 1/10 + 1/(-3000) ≈ -1/60; F = -60 cm.
Total internal reflection can occur at the interface between glass (n=1.5) and:
- A. Air (n=1.0)
- B. Water (n=1.33)
- C. Diamond (n=2.4)
- D. Glycerin (n=1.47)
Answer: A. Air (n=1.0)
Total internal reflection occurs when light travels from denser to rarer medium. Glass (1.5) to air (1.0) is denser to rarer, allowing TIR. Options b, c, d are rarer to denser or denser to denser.
Total internal reflection occurs when light travels from denser to rarer medium. Glass (1.5) to air (1.0) is denser to rarer, allowing TIR. Options b, c, d are rarer to denser or denser to denser.
An achromatic lens is made by combining a convex lens of crown glass (nᵣ=1.52, nᵥ=1.62) with a concave lens of flint glass (nᵣ=1.60, nᵥ=1.74). For achromatism, the condition is:
- A. f₁(nᵣ-1) + f₂(nᵣ-1) = 0
- B. f₁(nᵥ-nᵣ) + f₂(nᵥ-nᵣ) = 0
- C. f₁ + f₂ = 0
- D. Both b and c are correct
Answer: B. f₁(nᵥ-nᵣ) + f₂(nᵥ-nᵣ) = 0
For an achromatic doublet, the condition is f₁/ω₁ + f₂/ω₂ = 0, where ω is dispersive power (nᵥ-nᵣ)/(n-1). This simplifies to f₁(nᵥ-nᵣ) + f₂(nᵥ-nᵣ) = 0.
For an achromatic doublet, the condition is f₁/ω₁ + f₂/ω₂ = 0, where ω is dispersive power (nᵥ-nᵣ)/(n-1). This simplifies to f₁(nᵥ-nᵣ) + f₂(nᵥ-nᵣ) = 0.